It is known that f (x) = 2Sin (x + θ) 2）cos（x+θ 2）+2 3cos2（x+θ 2）- 3． (1) The analytic formula of F (x) is simplified; (2) If 0 ≤ θ≤ π, find θ to make the function f (x) even; (3) Under the condition of (2), the set of X satisfying f (x) = 1, X ∈ [- π, π] is obtained

It is known that f (x) = 2Sin (x + θ) 2）cos（x+θ 2）+2 3cos2（x+θ 2）- 3． (1) The analytic formula of F (x) is simplified; (2) If 0 ≤ θ≤ π, find θ to make the function f (x) even; (3) Under the condition of (2), the set of X satisfying f (x) = 1, X ∈ [- π, π] is obtained

（1）f（x）=sin（2x+θ）+2
3×1+cos(2x+θ)
2-
Three
=sin（2x+θ）+
3cos（2x+θ）
=2sin（2x+θ+π
3）；
(2) To make f (x) even, there must be f (- x) = f (x),
∴2sin（-2x+θ+π
3）=2sin（2x+θ+π
3) That is - sin [2x - (θ + π)
3）]=sin（2x+θ+π
3），
The results are as follows: - sin 2xcos (θ + π)
3）+cos2xsin（θ+π
3）=sin2xcos（θ+π
3）+cos2xsin（θ+π
3）
In other words, 2Sin 2xcos (θ + π)
3) = 0 holds for X ∈ R,
∴cos（θ+π
3) = 0, and 0 ≤ θ ≤ π,
Then θ = π
6；
(3) When θ = π
At 6, f (x) = 2Sin (2x + π)
2）=2cos2x=1，
∴cos2x=1
2，
∵x∈[-π，π]，
∴x=±π
6，
Then the set of X is {x | x = ± π
6}．

Given Tana = - 1 / 3, cos β = (√ 5) / 5, a, β∈ (0, π), find the maximum value of the function f (x) = √ 2Sin (x-a) + cos (x + β)

sinβ=(1-(1/5))^(1/2)=2(√5)/5
tana=-1/3<0 ,a∈(π/2,π)
sina/cosa=-1/3
(sina)^2/(1-(sina)^2)=1/9
sina=(√10)/10
cosa=-3(√10)/10
f(x)=√2sin(x-a)+cos(x+β)
=√2(sinx*cosa-cosx*sina)+(cosx*cosβ-sinx*sinβ)
=-(√5)sinx<=√5
Maximum = √ 5

P (COS α, sin α, 2Sin α) Q (2cos β, 2Sin β, 1) find the maximum and minimum of PQ How to calculate the extra - 4sin α + 4sin ^ 2 α

PQ = radical [(2cos β - cos α) ^ 2 + (2Sin β - sin α) ^ 2 + (1-2sin α) ^ 2]
=Radical [5-4cos (β - α) + (1-2sin α) ^ 2]]
α = - 90 degree, β = 90 degree, the maximum value of PQ is 3 pieces and 2 pieces
β = α = 30 degrees, PQ has a minimum value of 1

2. Find the maximum and minimum value of F (x) in the interval [- π / 6, π / 2]

F (x) = 2Sin (pai-x) cosx = 2sinxcosx = sin2x
Minimum positive period T = 2 π / 2 = π
-π/6

What are the maximum and minimum values of the function f (x) = [cos] squared x + 2Sin x

f(x)=1-(sinx)^2+2sinx
Let a = SiNx
Then - 1

The minimum positive period of the function y = 2Sin (PI / 3 minus x) minus cos (Pie / 6 + x) (x belongs to R) is urgent

Y = 2Sin (PAI / 3 minus x) minus cos (Pie / 6 + x)
=2cos[π/2-（π/3-x）]-cos(π/6+x)
=2cos(π/6+x)-cos(π/6+x)
=cos(π/6+x)
So the minimum positive period T = 2 π

F (x) = cos (2x - π / 3) + 2Sin (x - π / 4) sin (x + π / 4) is used to solve the equation of the minimum positive period and the symmetry axis of the image,

f（x）=cos（2x-π/3）+2sin（x-π/4）sin（x+π/4）= cos2x cosπ/3+ sin2x sinπ/3+2sin（x-π/4）cos（π/4-x）= cos2x cosπ/3+ sin2x sinπ/3+2sin（x-π/4）cos（x-π/4）= cos2x cosπ/3+ sin2x sinπ/3+ sin...

It was proved that - 2Sin α cos α + 1 / 1-2cos * 2 α = Tan α - 1 / Tan α + 1

Syndrome: (- 2Sin α cos α + 1) / (1-2cos? α) = (sin α - cos α) 2 / (sin? α - cos? α) = (sin α - cos α) 2 / [(sin α + cos α) (sin α - cos α)] = (sin α - cos α) / (sin α + cos α) = (Tan α - 1) / (Tan α + 1)

Conclusion: SiN4 α + Cos4 α = 1-2sin2 α Cos2 α

It is proved that the left = (sin2 α + Cos2 α) 2-2sin2 α Cos2 α = 1-2sin2 α Cos2 α = right,
Then sin 4 α + cos 4 α = 1-2 sin 2 α cos 2 α

The known function f (x) = 2 3sin(x+π 4)cos(x+π 4)-2sin(x+π)sin(x+5π 2) (I) find the minimum positive period and monotone increasing interval of F (x); (II) if the image of F (x) is shifted to the right π The graph of function g (x) is obtained by 12 units, and the function g (x) is found in the interval [0, π] 2] Maximum and minimum values on

（Ⅰ）f（x）=
3sin（2x+π
2）+2sinxcosx=
3cos2x+sin2x=2sin（2x+π
3），
∵ ω = 2, the minimum positive period of ᙽ f (x) is π;
Let 2K π - π
2≤2x+π
3≤2kπ+π
2, K ∈ Z, K π - 5 π
12≤x≤kπ+π
12，k∈Z，
Then the monotone increasing interval of F (x) is [K π - 5 π
12，kπ+π
12]，k∈Z；
(II) according to the meaning of the title: G (x) = 2Sin [2 (x - π)
12）+π
3]=2sin（2x+π
6），
∵2x+π
6∈[π
6，7π
6]，∴-1≤2sin（2x+π
6）≤2，
Then the maximum value of F (x) is 2 and the minimum value is - 1